Lemma. Let be analytic inside and on the boundary of the disk and satisfy for . If then for all in the disk.

Proof. Consider any circle . By the Cauchy Integral Formula we know that . For each on this circle with or . Assume that there is a on this circle for which . Then since is continuous there must be a whole arc of the circle on which for on . Also on the rest of the circle. Write .

From here how do we establish that , a contradiction? From the M-L Inequality, we know that . How do we get this to ?

Mar 12th 2009, 10:33 AM

ThePerfectHacker

Quote:

Originally Posted by manjohn12

Lemma. Let be analytic inside and on the boundary of the disk and satisfy for . If then for all in the disk.

This is similar to the maximum-modulos theorem. By Cauchy's theorem we know that:
Therefore, (by definition of contour integral)
Thus,

We have shown that on maximum value of on the disk is at least . However, and so the maximum value of must be equal to . Therefore, it follows that,
The only way that integral can equal to is it at each on we have .
Thus, for .
But was arbitrary and by shrinking and expanding we get the whole disk with .